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ES2C6 Week 5 Examples - QWCDWQE

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Electromechanical System Design (ES2C6)

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ES2C6 Week 5 Examples: Sensors

Lecture 5: Signal Conditioning

Question 1: A sensor has an output voltage range of 0mV to -500mV. The input voltage range of the

microprocessor used to measure the output is 0 to 10V. Design an op-amp based circuit to amplify

the output of the sensor to achieve a reasonable input voltage for the microcontroller.

Solution 1: The solution should use an inverting amplifier circuit with a gain of -20. The exact value

of the chosen resistors is not important, but their ratio is. However, the resistors should be roughly

between 1k and 500kΩ (the exact range will depend on the specific op-amp, but values in the region

of those given is typical).

2 1

200

20

10

out in

V R

V R

k

k

= −

= − = −

Question 2: A sensor has an output voltage range of 0mV to 300mV. The input voltage range of the

microprocessor used to measure the output is 0 to 5V. Design an op-amp based circuit to amplify the

output of the sensor to achieve a reasonable input voltage for the microcontroller.

Solution 2: The solution should use a non-inverting amplifier circuit with a gain of around 16 (16.

max). The exact value of the chosen resistors is not important as long as their values are reasonable,

but their ratio is important.

1 2

1

150

1 16

10

out in

V R

V R

k

k

= +

= + =

Question 3: Design a passive low pass filter with a cut-off frequency of 1 kHz.

Solution 3: This is a simple example, similar to the one given in the lecture. The equation for the cut-

off frequency is:

1

2

f c

RC

=

This can be rearranged to find a resistance or capacitance. In the lecture we chose a resistance and

found a capacitance. Generally, there is more choice of resistor values, so here we will choose a

capacitance and find a resistance:

1
2 c
R
f C
=

####### Choose 𝐶 = 10𝜇𝐹

( )( )

3 6

1
16
2 1 10 10 10
R
 −
=  
 

####### In fact, this is a large capacitor and a small resistor. A better design would be to pick a smaller

####### capacitor, perhaps 𝐶 = 10𝑝𝐹 which would give a resistor of 𝑅 ≈ 16𝑘Ω.

Lecture 5: Wheatstone Bridge

####### Question 4: The Wheatstone Bridge circuit given below gives an output reading of 54mV. What is

####### the value of R 1?

####### Solution 4: We derived the equation for R 1 in the lecture, so this is simply a case of inserting the

####### numbers given in the diagram and question.

( )

( )

( )

( ) ( )

3 4 1 2 3 4 4

1
12 30 10
30 1 87 k
0 30 10 12 10

S out S

V R R
R R
V R R V R
 + 
=  − 
 + + 
 
 + 
=  − = 
 + + 
 

####### Question 5: The Wheatstone bridge circuit in question four is part of a light sensor which is

####### connected to a microcontroller with an input range of 0 to 3 via a differential amplifier circuit,

####### Question 7: A Load Cell based on strain gauges and Wheatstone bridge circuits is used to measure a

####### mass. The measurement range of the load cell is 0 to 750kg, and its rated capacity is 750kg. The

####### rated output is 2 ± 0 𝑚𝑉 𝑉⁄. Give the range of output voltages if the excitation voltage is 10V

####### and the applied mass is 375kg. Assume that the behaviour of the sensor is 100% linear.

####### Solution 7: The given characteristics of the load cell mean that the output voltage at 750kg will

####### range from:

3 ,max 3 ,min

@ 750kg 2 10 10 20 mV
@ 750kg 1 10 10 19 mV

out out

V
V

− −

=   =
=   =

####### Because the behaviour of the sensor is linear, we can easily calculate the output for the 375kg mass:

375 kg 1
750 kg 2
=

,max ,max ,min ,min

1
@ 375kg @ 750kg 10 mV
2
1
@ 375kg @ 750kg 9 mV
2

out out out out

V V
V V
=  =
=  =

Lecture 5: Current Sensors

####### Question 8: A Hall effect current sensor has a Hall Coefficient of 𝐾𝐻 = 4 𝑚 3 𝐶−1, a magnetic flux

####### density of 𝐵 = 0 𝑇, and a plate thickness of 𝑡 = 2𝑚𝑚. What will the output voltage of the

####### sensor be if the current being measured is 𝐼 = 0𝐴.

####### Solution 8: Generally when using a hall effect current sensor the underlying equation and

####### parameters will not be given. Only the behaviour of the sensor such as its gain and operating range.

####### However, it is useful to have some understanding of the underlying process, which is what this

####### simple question explores. The equation is given in the lecture, and all that needs doing in inserting

####### the numbers:

3

0 0.
4 1 V
2 10

H

BI
V K
t −
= =  =

####### End of Questions

####### Mark Dooner, 2021

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ES2C6 Week 5 Examples - QWCDWQE

Module: Electromechanical System Design (ES2C6)

28 Documents
Students shared 28 documents in this course
Was this document helpful?
1
ES2C6 Week 5 Examples: Sensors
Lecture 5.2: Signal Conditioning
Question 1: A sensor has an output voltage range of 0mV to -500mV. The input voltage range of the
microprocessor used to measure the output is 0 to 10V. Design an op-amp based circuit to amplify
the output of the sensor to achieve a reasonable input voltage for the microcontroller.
Solution 1: The solution should use an inverting amplifier circuit with a gain of -20. The exact value
of the chosen resistors is not important, but their ratio is. However, the resistors should be roughly
between 1k and 500kΩ (the exact range will depend on the specific op-amp, but values in the region
of those given is typical).
2
1
200 20
10
out
in
VR
VR
k
k
=−
=− =−
Question 2: A sensor has an output voltage range of 0mV to 300mV. The input voltage range of the
microprocessor used to measure the output is 0 to 5V. Design an op-amp based circuit to amplify the
output of the sensor to achieve a reasonable input voltage for the microcontroller.
Solution 2: The solution should use a non-inverting amplifier circuit with a gain of around 16 (16.6
max). The exact value of the chosen resistors is not important as long as their values are reasonable,
but their ratio is important.
1
2
1
150
116
10
out
in
VR
VR
k
k
=+
=+ =
Question 3: Design a passive low pass filter with a cut-off frequency of 1 kHz.
Solution 3: This is a simple example, similar to the one given in the lecture. The equation for the cut-
off frequency is:
1
2
c
fRC
=
This can be rearranged to find a resistance or capacitance. In the lecture we chose a resistance and
found a capacitance. Generally, there is more choice of resistor values, so here we will choose a
capacitance and find a resistance: